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I have a data frame of about 500 rows and 170 columns. I am attempting to run a classification model with svm from the e1071 package. The classification variable is called 'SEGMENT', a factor variable with 6 levels. There are three other factor variables in the data frame, and the rest are numeric.

data <- my.data.frame
# Split into training and testing sets, training.data and testing.data
.
.
.
fit <- svm(SEGMENT ~ ., data = training.data, cost = 1, kernel = 'linear', 
+ probability = T, type = 'C-classification')

The model runs fine.

Parameters:
SVM-Type:  C-classification 
SVM-Kernel:  linear 
   cost:  1 
   gamma:  0.0016 

Number of Support Vectors:  77

( 43 2 19 2 2 9 )

Number of Classes:  6 

Levels: 
EE JJ LL RR SS WW

The problem arises when I try to test the model on data.testing, which is structured exactly like the training set:

x <- predict(fit, testing.data, decision.values = T, probability = T)

And then things blow up rather spectacularly:

Error in predict.svm(fit, newdata = testing, decision.values = T, probability = T) : 
test data does not match model !

Ideas are most welcome.

Dino Fire
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    Post output from `str(testing.data)`. My guess is that the factor levels will be different. I'm also guessing that if you search on that error text in SO htat you will find this has been asked and answered several times before. – IRTFM Oct 08 '14 at 19:57
  • Please take the time to post a [reproducible example](http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) to make it easier to help you. Otherwise we are merely guessing and what might be wrong. – MrFlick Oct 08 '14 at 20:06

1 Answers1

4

This happens when the columns in test and train data aren't same. Try str(training.data) & str(testing.data) they should have the same variables except for the one that needs to be predicted. Include only those factors you want to use for prediction in the svm training model.

For eg:

fit <- svm(SEGMENT ~ ., data = training.data[,1:6], cost = 1, kernel = 'linear', 
+ probability = T, type = 'C-classification')     


x <- predict(fit, testing.data[,1:5], decision.values = T, probability = T)
Abhinav Pathak
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